Let $E_1, E_2$ and $E_3$ be the events of a driver being a scooter driver, car driver and truck driver respectively. Let A be the event that the person meets with an accident. There are 2000 insured scooter drivers, 4000 insured car drivers and 6000 insured truck drivers. Total number of insured vehicle drivers = 2000 + 4000 + 6000 = 12000 ∴ $P(E_1)$ = $\frac{2000}{12000}$ = $\frac{1}{6}$ , $P(E_2)$ = $\frac{4000}{12000}$ = $\frac{1}{3}$ , $P(E_3)$ = $\frac{6000}{12000}$ = $\frac{1}{2}$ Also, we have: P (A|$E_1$) = 0.01 = $\frac{1}{100}$ P (A|$E_2$) = 0.03 = $\frac{3}{100}$ P (A|$E_3$) = 0.15 = $\frac{15}{100}$ Now, the probability that the insured person who meets with an accident is a scooter driver is P (A|$E_1$). Using Bayes’ theorem, we obtain: P ($E_1$|A) = = $\frac{\frac{1}{6} \times \frac{1}{100}}{\frac{1}{6} \times \frac{1}{100} + \frac{1}{3} \times \frac{3}{100} + \frac{1}{2} \times \frac{15}{100}}$ = $\frac{\frac{1}{6}}{\frac{1}{6} + 1 + \frac{15}{2}}$ = $\frac{1}{6} \times \frac{6}{52}$ = $\frac{1}{52}$